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Evaluate (√3 + √2)⁶ - (√3 - √2)⁶

Solution:

Using binomial theorem, we will find the expansions of (√3 + √2)⁶ and (√3 - √2)⁶.

Then we get, (√3 + √2)6 - (√3 - √2)6 

= [6C₀ (√3)6 + 6C₁ (√3)5 (√2)1 + 6C₂ (√3)4 (√2)2 + 6C₃ (√3)3 (√2)3 

+ 6C₄ (√3)2 (√2)4 + 6C₅ (√3)1 (√2)5 + 6C₆ (√2)6]
- [6C₀ (√3)6 - 6C₁ (√3)5 (√2)1 + 6C₂ (√3)4 (√2)2 - 

6C₃ (√3)3 (√2)3 + 6C₄ (√3)2 (√2)4 - 6C₅ (√3)1 (√2)5 + 6C₆ (√2)6]

= 2 [6C₁ (√3)5 (√2)1 + 6C₃ (√3)3 (√2)3 + 6C₅ (√3)1 (√2)5]

Using nCr formula,

= 2 [ 6 (9√3) (√2) + 20(3√3)(2√2) + 6(√3)(4√2)]

= 2 [54√6 + 120√6 + 24√6]

= 2 [ 198 √6]

= 396 √6

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 5

Evaluate (√3 + √2)⁶ - (√3 - √2)⁶

Summary:

We evaluated the value of (√3 + √2)⁶ - (√3 - √2)⁶ to be 396√6

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